dynamically populate mysql array into drop down menu - php

I have a database that I want to get data out onto a website. It contains states listed by name and id. Counties listed by id, namne , and state that contains thems ID and then clubs that exist , with a reference to the county id's that they exist in and columns for their actual data.
What I've got :
A drop down menu that populates itself with state id and name.
What I'd like to accomplish:
On selection of state , let's say ny , take it's id and use this in gathering another mysql array for the county drop down. I'd like it to dynamically occur on selection of state , maybe even giving a count of results next to the drop down.
$resstate = mysql_query("SELECT * FROM state ORDER by longstate;") or die("Note: " . mysql_error());
State:
<select name="State" size=1>
<?
while( $rs = mysql_fetch_array( $resstate ) ) {
echo "<option value=" .$rs['id'] . ">" . $rs['longstate'] . "</option>";
}
echo "</select>";
?>
I know I could use a JavaScript onChange="this.form.submit()" on the first drop down, but it's my understanding that I'd then be making a new page at that point and don't know if I could keep the functionality of the state drop down, say if you accidentally chose new Hampshire when you wanted New York.
here's an example of the current array filling the drop down :
http://snowmobileamerica.com/countytest.php
----EDIT---
Using Dagons Advice , I looked into Ajax.
I made a php file that's supposed to query the database based on a reference to getcounty.php?q=
The file is created as follows :
<?php
$q=$_GET["q"];
$cn=mysql_connect("localhost","user","password") or die("Note: " . mysql_error());
mysql_select_db("snowusa_clubs", $cn);
$sql="SELECT * FROM county WHERE state_id = '".$q."' ORDER by name";
$result = mysql_query($sql);
echo "<select name="County" size=1>";
while($rc = mysql_fetch_array($result))
{
echo "<option value=" .$rc['id'] . ">" . $rc['name'] . "</option>";
}
echo "</select>";
mysql_close($cn);
?>
If i try to run it manually http://www.snowmobileamerica.com/getcounty.php?q=33 I get a 500 internal server error...
Any ideas where I went wrong?

try adding an id to the element, then make an ajax call to a handler with jquery:
$("#State").change(function() {
$.post("path/to/request handler/" , { "State" : $(this).val() },
function(data){
if (data == "OK"){
//add some elements here
} else {
//handle an error here
}
});
});

not able to comment yet.
but for the second question try:
<?php
$q=$_GET["q"];
$cn=mysql_connect("localhost","user","password") or die("Note: " . mysql_error());
echo "Conn ok<br>";
mysql_select_db("snowusa_clubs", $cn);
echo " Database opened<br>";
$sql="SELECT * FROM county WHERE state_id = '$q' ORDER by name";
$result = mysql_query($sql);
echo " Database queried <br>";
echo "<select name='County' size=1>";
while($rc = mysql_fetch_array($result))
{
echo "<option value='" .$rc['id'] . "'>" . $rc['name'] . "</option>";//added single quotes in the value
}
echo "</select> ";
mysql_close($cn);
?>

Related

Changing the output of my HTML form

I am trying to create a form using data from MySQL and my current output is not looking as expected so I just need some simple changes to my current form
My current output in the browser: http://gyazo.com/f4668ca59586ec0d4d1500ea6f7b6257
What I am looking to do is
fix the problem with the first column called 'Long Jump A' so its
next to a drop down menu, and you can see the bottom one is next to
nothing.
Secondly, How can I swap the order of the drop down and the
event? So the event is on the left then the drop down is on the
right?
Moving the form down from the top so there is a space
The code:
<?php
require_once 'db/connect.php';
//Query to display all events
if ($event_result = $con->query("SELECT Event.Name FROM event")) {
echo "<form method =\"POST\">";
while ($row = $event_result->fetch_assoc()) {
echo $row['Name'] . ' <br> ';
if ($student_result = $con->query("SELECT Student.Form, Teacher.Form, Student.Forename, Student.Surname, Student_ID " .
"FROM Student, Teacher " .
"WHERE Student.Form = Teacher.Form AND Teacher.Form = 'C'")) {
if ($student_result->num_rows) {
echo "<select name ='Student_ID'>";
while ($row1 = $student_result->fetch_assoc()) {
echo "<option value ='" . $row1['Student_ID'] . "'>" . $row1['Forename'] . ' ' . $row1['Surname'] . "</option>";
}
echo "</select>";
}
}
}
echo "</form>";
}
?>
You have an opening <br /> as you echo out the name of the row. This means echo the title, then break before you display the <select> group. You should move that <br /> down right before the closing while loop.
Or just remove the top <Br /> and add it back after the ending so it looks like this:
echo "</select> <br />";
edit: this will fix the last two of your questions, as now each select group will properly line up with the title of the row. If you want more space at the top of the form you have a bunch of options: add more breaks, add a margin or padding, or position the element as relative and use css top property or CSS transforms to move it.

How do I dynamically display the results of my PHP results in separate divs according to their ID?

I'm trying to create a simple e-commerce system. First thing I did was select the rows with the same Order ID from mysql. My table looks like this:
Now, I'd like to know how I can group them into separate divs, it should look like this:
Here's my code:
$con = mysqli_connect("localhost","root","","test");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result2 = mysqli_query($con, "SELECT DISTINCT request_date, department_id FROM tbl_requests WHERE request_id=".$_GET['request_id']) ;
while($row2 = mysqli_fetch_array($result2)) {
echo "" . $row2['request_date'] . "<br/>";
echo "Order from Department " . $row2['department_id'] . "<br/>";
echo "<br/>";
echo "<hr/>";
echo "<br/>";
}
$result = mysqli_query($con,"SELECT * FROM tbl_requests WHERE request_id=".$_GET['request_id']);
while($row = mysqli_fetch_array($result)) {
echo "" . $row['request_details'] . "";
echo "<br/>";
}
I'm sorry if ever this question is incomplete, please feel free to ask me any more questions. Thank you in advance :)
You can check for every product in array using Javascript or jQuery(much easier).
This way you can check if your page does contain any existing div with that manufacture id (ie. #m_1055, #m_1040) or not.
If div does exist, append product in that div (in jQuery.append())
If not, then first append the div with that manufacture id to the document and then append the product to it.

Passing selected variables in PHP?

I want to send some information from one php file to another.
I've read about the use of $_SESSION and $_POST, but they're giving me some problems.
My code looks something like this:
<form action="booking.php" method="post">
<select name="bookingflight">
<?php
$query = "SELECT Flight, Name
FROM Airport";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)) {
echo "<option value=\"". $row['Flight'] . ',' . $row['name'] . "\">" . $row['Flight'] . ' ' . $row['name'] . "</option>";
}
?>
</select></p>
<input type="submit" value="book flight"/>
</form>
This gives me a dropbox list of all the flights, along with the name of the flights.
If I select an element it's stored in $_POST["bookingflight"] which I can access in booking.php, which is fine.
However, it's given as a string, while I should be able to handle flight and name separately.
Ideally, I'd have two variables, one for flight and one for name, which I can access in booking.php.
How should I do this? With $_SESSION I don't even know how to assign a selected item from the list to a variable.
Alternate Solution: If Flight is unique data
while($row = mysql_fetch_array($result)) {
$flight=$row['Flight'];
$name=$row['Name'];
echo "<option value='$flight'>$flight $name</option>";
}
PHP : booking.php
<?php
$flight=$_POST['bookingflight'];
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT Flight,Name FROM Airport WHERE Flight='$flight'");
$row = mysqli_fetch_array($result);
$name= $row['Name'] ; // Here you get Name of selected Flight
mysqli_close($con);
?>
Or
If your Airport table contains any unique data/id , you can pass that data as option value too
You started this the wrong way...
Give your Airport table a unique primary key (named id, INT, auto-increment) and pass that as a value in generated options:
echo "<option value=\"". $row['id'] ."\">" . $row['Flight'] . ' ' . $row['name'] . "</option>";
Now when you POST your form you get that ID value in booking.php.
Because every flight has a unique ID you can just issue another query and you get your result as an array there:
$query = "SELECT Flight, Name FROM Airport WHERE id = $id";
Put this in your booking.php...
if($_POST)
{
$values = $_POST['bookingflight'];
$val= explode(',',$values);
$_SESSION['flightnumber'] = $val[0];
$_SESSION['flightname'] = $val[1];
}

How do I create a dynamic set of pull down menus using javascript

I am working on optimizing a form for our website that lets a user complete the it using pull down menus. I have figured out how to populate a single pull down menu based on a PHP query, but I then need another pull down menu to be populated based on that selection. I am attempting to use the onchange attribute for the select tag. I know that my query in the javascript is not dynamic yet, but I am just trying to get it to show the next pull down first, which it isn't. When I select something from the first pull down nothing happens. Thank you advance for any help.
$query2 = "select distinct Provider_Name from CE_ACTIVITY_LIST_T where IS_ACTIVE = 'YES'";
$result2 = mysql_query($query2, $link) or die(mysql_error());
//$row2 = mysql_fetch_array($result2) or die(mysql_error());
echo "<table border='0'><tr><th>";
echo "Course Provider:</th><td><select name='providerName' id='provider'onchange='getResult()'>";
echo "<option SELECTED>Pick Provider</option>";
$i=0;
while($row2 = mysql_fetch_array($result2))
{
echo "<option value=\"" . $row2['Provider_Name'] . "\">" . $row2['Provider_Name'] . "</option>";
$i++;
}
echo "</select></td></tr><tr></tr>";
<script type="text/javascript">
function getResult(field)
{
var field=field;
var isEqualTo=document.getElementById('provider');
document.getElementById('selectCourse').innerHTML = '<?php
include ('connectionOpen.php');
$queryCourse="select Course_Title from CE_ACTIVITY_LIST_T where Provider_Name = 'Test 1' and isActive = 'YES'";
echo "<tr><th>";
echo "Course Title:</th><td><select name='courseTitle'>";
echo "<option SELECTED>Which course?</option>";
$i=0;
while($row2 = mysql_fetch_array($result2))
{
echo "<option value=\"" . $row2['Course_Title'] . "\">" . $row2['Course_Title'] . "</option>";
$i++;
}
echo "</select>";
echo "</td></tr></table>";
?>';
}
</script>
If you want to populate another list then you need to load it with ajax, you can use jQuery to make it easier. First create php script that will generate html for second option list, eg. myscript.php, this script should receive parameter name.
Be sure that jQuery is included. Then modify your javascript as follows:
function onChange()
{
var name = providerName.value;
var url = "path-to-script";
$.get(url, {name:name} function(html){
$("#result").html(html)
});
}
Instead of second option use <div id='result'></div> as placeholder for new option list.

PHP Dropdown of all records from database + select current set record

I have a PHP dropdown of a list of groupnames (together with id, so it can be updated). In this FORM page you can change the groupname specified for an item by choosing possibilities from the dropdown coming out from the database. My code below works, but there must be a better way, because I get the first field as the currently set, and then all the possibilities, so I get this record twice.
Example:
- Keyboard (Currently set)
- Speakers (Possible to choose, straight from DBS)
- Midi Controllers (Possible to choose, straight from DBS)
- Keyboard (Possible to choose, straight from DBS)
- Drum set (Possible to choose, straight from DBS)
As you see I get the currently set record again.
My code:
echo "<select name='itemgroupid'>";
// CHOOSE CURRENT SET RECORD AS SELECTED ITEM
echo "<option value='" . $itemgroupid . "'>";
$selected="
SELECT item.itemid, itemgroup.itemgroupname, itemgroup.itemgroupid
FROM item, itemgroup
WHERE item.itemid=$itemid";
$selectedresult=mysql_query($query) or die("query fout " . mysql_error() );
while($record=mysql_fetch_array($selectedresult) ) {
echo "" . $itemgroupname . "</option>";
}
// QUERY TO SHOW ALL POSSIBLE CHOOSABLE RECORDS FROM DATABASE
$itemgroupquery="SELECT itemgroupname,itemgroupid FROM itemgroup";
$itemgroupqueryresult = mysql_query ($itemgroupquery);
while($nt=mysql_fetch_array($itemgroupqueryresult)){
echo "<option value=$nt[itemgroupid]>$nt[itemgroupname]</option>";
}
echo "</select>";
There are 2 ways to achieve what you're looking for:
1) To show the selected item at the top of the dropdown
echo "<select name='itemgroupid'>";
// CHOOSE CURRENT SET RECORD AS SELECTED ITEM
echo "<option value='" . $itemgroupid . "'>";
$selected="
SELECT item.itemid, itemgroup.itemgroupname, itemgroup.itemgroupid
FROM item, itemgroup
WHERE item.itemid=$itemid";
$selectedresult=mysql_query($query) or die("query fout " . mysql_error() );
while($record=mysql_fetch_array($selectedresult) ) {
echo "" . $itemgroupname . "</option>";
}
// QUERY TO SHOW ALL POSSIBLE CHOOSABLE RECORDS FROM DATABASE
$itemgroupquery="SELECT itemgroupname,itemgroupid FROM itemgroup WHERE item.itemid != $itemid";
$itemgroupqueryresult = mysql_query ($itemgroupquery);
while($nt=mysql_fetch_array($itemgroupqueryresult)){
echo "<option value=$nt[itemgroupid]>$nt[itemgroupname]</option>";
}
echo "</select>";
2) Show the selected item in it natural place
echo "<select name='itemgroupid'>";
// QUERY TO SHOW ALL POSSIBLE CHOOSABLE RECORDS FROM DATABASE
$itemgroupquery="SELECT itemgroupname,itemgroupid FROM itemgroup";
$itemgroupqueryresult = mysql_query ($itemgroupquery);
while($nt=mysql_fetch_array($itemgroupqueryresult)){
echo "<option value=$nt[itemgroupid]";
if( $itemid == $nt['itemgroupid'] ) echo ' selected="selected"';
echo ">$nt[itemgroupname]</option>";
}
echo "</select>";
HTH
OK
In your code.
rather than output your selected value at the top, do it the proper way :)
Select your current item (make a note of the itemgroupid for example)
then in your output
while($nt=mysql_fetch_array($itemgroupqueryresult)){
echo "<option ";
if ($savedid==$nt[itemgroupid]) echo "selected ";
echo "$nt[itemgroupid]>$nt[itemgroupname]</option>";
}
echo "</select>";
This will produce with $savedid=1
<option value=0>group 0</option>
<option selected value=1>group 1</option>
<option value=2>group 2</option>
Add the default selected record to a empty array first like
toDisplay = array('selected_record');
Then, get the data from the database using your sql and append it to this array.
Later run a array_unique on it and finally using a loop create the html output string, in the same way you are doing it now.

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